Ta có:

7A=7*(7^2010+1)/7^2011-1=7^2011+7/7^2011-1=(7^2011-1)+8/7^2011-1=1+8/7^2011-1

7B=7*(7^2011+1)/7^2012-1=7^2012+7/7^2012-1=(7^2012-1)+8/7^2012-1=1+8/7^2012-1

Vì 8/7^2011-1>8/7^2012-1 nên 1+8/7^2011-1>1+8/7^2012-1 hay A>B

Vậy: A>B

đúng cho mình nha ! . . .=) . . .

A > B

tk nha

ta có:\(7A=\frac{7\left(7^{2010}+1\right)}{7^{2011}-1}=\frac{7^{2011}+7}{7^{2011}-1}=\frac{7^{2011}-1+8}{7^{2011}-1}=\frac{7^{2011}-1}{7^{2011}-1}+\frac{8}{7^{2011}-1}=1+\frac{8}{7^{2011}-1}\)

\(7B=\frac{7\left(7^{2011}+1\right)}{7^{2012}-1}=\frac{7^{2012}+7}{7^{2012}-1}=\frac{7^{2012}-1+8}{7^{2012}-1}=\frac{7^{2012}-1}{7^{2012}-1}+\frac{8}{7^{2012}-1}=1+\frac{8}{7^{2012}-1}\)

vì 7²⁰¹¹-1<7²⁰¹²-1

\(\Rightarrow\frac{8}{7^{2011}-1}>\frac{8}{7^{2012}-1}\)

=>A>B

Ta có:2012/2010=1+1/2011+1/2012 (1)

Thay (1) vào A, ta có:

2010/2011+2011/2012+1+1/2011+1/2012

= (2010/2011+1/2011) + 1+ (2011/2012+1/2012)

=1+1+1=3=51/17

suy ra A>51/17

Ta có B=1/3+1/4+...+1/17(có 15 sh)

B=(1/17+1/3).15:2

B=50/17(2)

Từ (1) và (2)=>A>B

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)

\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)

\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)

\(=\frac{1}{3}+\frac{1}{4}\)

\(=\frac{7}{12}\)

\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)